Problem 54 - Algebra and Trigonometry

Question

On a sheet of graph paper, or using a graphing calculator, draw the parabola \(y=x^{2} .\) Then draw the graphs of the linear equation \(y=x+k\) on the same coordinate plane for various values of \(k .\) Try to choose values of \(k\) so that the line and the parabola intersect at two points for some of your \(k^{\prime} \mathrm{s}\) , and not for others. For what value of \(k\) is there exactly one intersection point? Use the results of \(k\) is there exactly one make a conjecture about the values of \(k\) for which the following system has two solutions, one solution, and no solution. Prove your conjecture. $$\left\\{\begin{array}{l}{y=x^{2}} \\ {y=x+k}\end{array}\right.$$

Step-by-Step Solution

Answer

The parabola and line intersect at exactly one point when \(k = -\frac{1}{4}\).

Step 1: Graph the Parabola

Start by sketching the graph of the parabola given by the equation \(y = x^2\). This is a standard parabola opening upwards with its vertex at the origin (0,0) on the coordinate plane.

Step 2: Graph Linear Equations for Various k

Draw the graph of the linear equation \(y = x + k\) for several values of \(k\). Plot these lines on the same graph as the parabola. Observe where the line intersects the parabola for different values of \(k\). For example, try \(k = -2, 0, 1, 2\).

Step 3: Analyze Intersections

Observe the points where each linear graph intersects the parabola. For some values of \(k\), the line will intersect the parabola at two distinct points, at one point (tangent), or not at all.

Step 4: Find the Value of k for One Intersection

To find the value of \(k\) for which the line intersects the parabola at exactly one point, set the equations equal to each other: \(x^2 = x + k\). Rearrange into the standard quadratic form: \(x^2 - x - k = 0\). The line is tangent to the parabola when the quadratic equation has exactly one solution, which occurs when the discriminant \(b^2 - 4ac\) is zero. Here, \(a = 1\), \(b = -1\), and \(c = -k\).

Step 5: Solve for the Discriminant

Calculate the discriminant of the quadratic equation: \((-1)^2 - 4(1)(-k) = 1 + 4k\). Set this equal to zero to find when the line is tangent: \[1 + 4k = 0\]Solve for \(k\): \[k = -\frac{1}{4}\]

Step 6: Conjecture and Prove the Number of Solutions

Conjecture based on the discriminant's analysis:- Two solutions: \(1 + 4k > 0\) which implies \(k > -\frac{1}{4}\).- One solution: \(1 + 4k = 0\) which implies \(k = -\frac{1}{4}\).- No solution: \(1 + 4k < 0\) which implies \(k < -\frac{1}{4}\). You can verify these conditions using additional test values for \(k\) and observing the intersection behavior of the line and parabola.

Key Concepts

Parabola

A parabola is a U-shaped curve that you can represent using a quadratic equation. Quadratic equations like to take the form of \(y = ax^2 + bx + c\). In the simplest case, you encounter the parabola with the equation \(y = x^2\), which opens upwards. The vertex, or the lowest point of this parabola, is at the origin (0, 0).
- **Vertex**: The vertex is the point where the parabola changes direction. In our basic example \(y = x^2\), it is (0, 0).
- **Axis of Symmetry**: The parabola is symmetric along a vertical line known as the axis of symmetry. Here, it is the y-axis or \(x = 0\).
When you sketch a parabola like \(y = x^2\), you can see it bend elegantly upwards, creating a symmetrical, distinctive shape.

Linear Equations

Linear equations form straight lines when graphed and have the general form \(y = mx + b\), where \(m\) represents the slope and \(b\) is the y-intercept. For our case, the equation \(y = x + k\) describes a family of lines that all have a slope of 1 and adjust vertically according to the value of \(k\).
- **Slope (m)**: The slope tells us how steep the line is. Here, the slope is 1, indicating that for each unit increase in \(x\), \(y\) increases by the same amount.
- **Y-intercept (b)**: This is the point where the line crosses the y-axis. In \(y = x + k\), the \(k\) value shifts the entire line up or down.
By changing \(k\), you shift the line vertically while maintaining its direction and angle.

Intersection Points

An intersection point is where two graphs meet. When analyzing the intersection between a parabola and a line, we try to find common solutions to their equations.
For our quadratic \(y = x^2\) and linear \(y = x + k\) equations, setting them equal reveals potential intersection points: \(x^2 = x + k\). Solving this equation determines the \(x\) values where both functions have the same \(y\).
- **Two Intersection Points**: This happens when the line crosses the parabola, which usually gives you two distinct \(x\) values.
- **One Intersection Point**: Occurs when the line is tangent to the parabola, touching it in exactly one spot.
- **No Intersection Points**: Means that the line does not touch or cross the parabola at all.
The nature of solutions depends on the geometry and alignment of these two functions.

Discriminant

The discriminant helps determine the nature of the roots of a quadratic equation. It is part of the quadratic formula \( ax^2 + bx + c = 0 \) and is given by \(b^2 - 4ac\).
In the context of our function \(x^2 = x + k\), rearranging gives \(x^2 - x - k = 0\). Here, the discriminant is \(1 + 4k\), derived from \((-1)^2 - 4(1)(-k)\).
- **Positive Discriminant**: \(1 + 4k > 0\) implies two distinct real solutions, meaning two intersection points.
- **Zero Discriminant**: \(1 + 4k = 0\) implies exactly one real solution, creating a tangent point. Solving this gives \(k = -\frac{1}{4}\).
- **Negative Discriminant**: \(1 + 4k < 0\) results in no real solution, meaning no intersection.
This mathematical tool is crucial for predicting how a line will interact with a parabola graphically.